1956 AHSME Problems/Problem 15
Revision as of 19:07, 12 February 2021 by Coolmath34 (talk | contribs) (Created page with "== Problem == The root(s) of <math>\frac {15}{x^2 - 4} - \frac {2}{x - 2} = 1</math> is (are): <math>\textbf{(A)}\ -5\text{ and }3\qquad \textbf{(B)}\ \pm 2\qquad \textbf{...")
Problem
The root(s) of 
is (are):
Solution
Use a common denominator: 
. After moving the 
to the left side, notice that you can factor the numerator:
![\[\frac{(x-5)(x+3)}{x^2-4} = 0\]](http://latex.artofproblemsolving.com/c/a/8/ca8127a750c9d9a3fca0ee31600cd40c9e563c3e.png)
The roots of this equation are 
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 14 |
Followed by Problem 16 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
